022 Exam 2 Sample A, Problem 3

Find the antiderivative of $\int {\frac {1}{3x+2}}\,dx.$

ExpandFoundations:
This problem requires two rules of integration. In particular, you need
Integration by substitution (u - sub): If $u=g(x)$  is a differentiable functions whose range is in the domain of $f$, then
$\int g'(x)f(g(x))dx\,=\,\int f(u)du.$
We also need the derivative of the natural log since we will recover natural log from integration:
$\left(ln(x)\right)'\,=\,{\frac {1}{x}}$

 Solution:

ExpandStep 1:
Use a u-substitution with $u=3x+2.$ This means $du=3\,dx$, or $dx=du/3$. After substitution we have $\int {\frac {1}{3x+2}}\,=\,\int {\frac {1}{u}}\,{\frac {du}{3}}\,=\,{\frac {1}{3}}\int {\frac {1}{u}}\,du.$

ExpandStep 2:
We can now take the integral remembering the special rule:
${\frac {1}{3}}\int {\frac {1}{u}}\,du\,=\,{\frac {\log(u)}{3}}.$

ExpandStep 3:
Now we need to substitute back into our original variables using our original substitution $u=3x+2$
to find  ${\frac {\log(u)}{3}}={\frac {\log(3x+2)}{3}}.$

ExpandStep 4:
Since this integral is an indefinite integral we have to remember to add a constant  $C$ at the end.

ExpandFinal Answer:
$\int {\frac {1}{3x+2}}\,dx\,=\,{\frac {\ln(3x+2)}{3}}+C.$

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